Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $t \neq 0$. $q = \dfrac{-4t^2 + 24t}{5t^3 - 40t^2 + 60t} \div \dfrac{2t + 20}{t^2 + t - 90} $
Explanation: Dividing by an expression is the same as multiplying by its inverse. $q = \dfrac{-4t^2 + 24t}{5t^3 - 40t^2 + 60t} \times \dfrac{t^2 + t - 90}{2t + 20} $ First factor out any common factors. $q = \dfrac{-4t(t - 6)}{5t(t^2 - 8t + 12)} \times \dfrac{t^2 + t - 90}{2(t + 10)} $ Then factor the quadratic expressions. $q = \dfrac {-4t(t - 6)} {5t(t - 6)(t - 2)} \times \dfrac {(t + 10)(t - 9)} {2(t + 10)} $ Then multiply the two numerators and multiply the two denominators. $q = \dfrac {-4t(t - 6) \times (t + 10)(t - 9) } { 5t(t - 6)(t - 2) \times 2(t + 10)} $ $q = \dfrac {-4t(t + 10)(t - 9)(t - 6)} {10t(t - 6)(t - 2)(t + 10)} $ Notice that $(t - 6)$ and $(t + 10)$ appear in both the numerator and denominator so we can cancel them. $q = \dfrac {-4t(t + 10)(t - 9)\cancel{(t - 6)}} {10t\cancel{(t - 6)}(t - 2)(t + 10)} $ We are dividing by $t - 6$ , so $t - 6 \neq 0$ Therefore, $t \neq 6$ $q = \dfrac {-4t\cancel{(t + 10)}(t - 9)\cancel{(t - 6)}} {10t\cancel{(t - 6)}(t - 2)\cancel{(t + 10)}} $ We are dividing by $t + 10$ , so $t + 10 \neq 0$ Therefore, $t \neq -10$ $q = \dfrac {-4t(t - 9)} {10t(t - 2)} $ $ q = \dfrac{-2(t - 9)}{5(t - 2)}; t \neq 6; t \neq -10 $